Problem: Let $y=\cos\left(4x-\dfrac{\pi}{3}\right)$. What is the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12$ (Choice B) B $2$ (Choice C) C $-\dfrac{\sqrt{3}}{2}$ (Choice D) D $-2\sqrt{3}$
Let's start by finding the expression for $\dfrac{dy}{dx}$. Then, we can evaluate it at $x=\dfrac{\pi}{6}$. $\cos\left(4x-\dfrac{\pi}{3}\right)$ is a trigonometric expression, but its argument isn't simply $x$. Therefore, it defines a composite trigonometric function. In other words, suppose $u(x)=4x-\dfrac{\pi}{3}$, then $y=\cos\Bigl(u(x)\Bigr)$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[\cos\Bigl(u(x)\Bigr)\right]=-{\sin\Bigl(u(x)\Bigr)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\cos\left(4x-\dfrac{\pi}{3}\right) \\\\ &=\dfrac{d}{dx}\cos\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=4x-\dfrac{\pi}{3}} \\\\ &=-{\sin\Bigl(u(x)\Bigr)}\cdot u'(x) \\\\ &=-{\sin\Bigl(4x-\dfrac{\pi}{3}\Bigr)}\cdot 4&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $x={\dfrac{\pi}{6}} $. $\begin{aligned} &\phantom{=}-{\sin\Bigl(4\left({\dfrac{\pi}{6}}\right)-\dfrac{\pi}{3}\Bigr)}\cdot 4 \\\\ &=-4\sin\left(\dfrac{\pi}{3}\right) \\\\ &=-4\cdot \dfrac{\sqrt{3}}{2} \\\\ &=-2\sqrt{3} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{6}$ is $-2\sqrt{3}$.